Learned from Alcumus

Volume of square pyramid

1/3 * area of base * height

Inscribed angles

= 1/2 of the angle of the arc they subtend

altitude of a triangle

A line from a vertex to the opposite side of the triangle, perpendicular to that side. Does not always end at the midpoint of that side, but does in the case of the incongruent side of an isosceles triangle. This is also the only case where the angle is bisected by the altitude. The altitude from side a (Ha) (in terms of side lengths and the semiperimeter) = 2*sqrt(s*(s-a)(s-b)(s-c))/a

An ice cream scoop scoops out ice cream spheres with radius 1 inch. How many scoops are needed to fill an ice cream cone with radius 2 inches and height 5 inches?

Each ice cream sphere has volume 4/3pi (1^3) = 4/3pi cubic inches. The ice cream cone holds 1/3pi (2^2)(5) = 20/3pi cubic inches. 20/3pi/4/3pi = 5, so we need {5} scoops to fill the cone.

Angle bisector theorem

If a line CX bisects an angle ACB, then AC/AX = BC/BX

Angle Bisector theorem

If an angle in a triangle ABC is bisected by a line that runs to point X on the opposite side starting from point C, then AX/AC = BX/BC

In how many different ways can four students stand in a straight line if two of the students refuse to stand next to each other?

We will count the number of ways in which the two students do stand next to each other and then subtract that from the total number of ways all four students can stand in a line without the restriction. If the two students stand next to each other, then we can treat them as a block. There are three blocks: two one-student blocks and one two-student block. We can arrange the blocks in 3! = 6 ways, and there are 2 ways to arrange the students within the two-student block, for a total of 6*2 = 12 ways. The total number of ways to arrange all four students in a line without restrictions is 4!=24 ways. Thus, the number of ways with the restriction is 24-12={12} ways.

The image of the point with coordinates $(-3,-1)$ under the reflection across the line $y=mx+b$ is the point with coordinates $(5,3)$. Find $m+b$.

The line of reflection is the perpendicular bisector of the segment connecting the point with its image under the reflection. The slope of the segment is 3--1/5--3=1/2 Since the line of reflection is perpendicular, its slope, $m$, equals $-2$. By the midpoint formula, the coordinates of the midpoint of the segment is ({5-3}/2,{3-1}/2))=(1,1). Since the line of reflection goes through this point, we have $1=(-2)(1)+b$, and so $b=3$. Thus $m+b={1}.$

Exterior angle theorem

The measure of an exterior angle (our w) of a triangle equals to the sum of the measures of the two remote interior angles (our x and y) of the triangle.

What is the mean of the measures of the three exterior angles of a triangle if two of the interior angles have measures of 63 and 78 degrees?

The measure of an exterior angle (our w) of a triangle equals to the sum of the measures of the two remote interior angles (our x and y) of the triangle. But also, The exterior angles of a triangle sum to 360º, so the average of their measures is $360º/3 = 120º.

In how many ways can 8 people be seated in a row of chairs if three of the people, John, Wilma and Paul, refuse to sit in three consecutive seats?

The number of all seating arrangements is $8!$. The number of seating arrangements in which John, Wilma and Paul sit next to each other is $6!\times 3!$. We can arrive at $6!\times 3!$ by considering John, Wilma and Paul as one person, arranging the ``six'' people (the JWP super-person plus the 5 other people) first, then arranging John, Wilma and Paul. Thus the number of acceptable arrangements is 8!-(6!*3!) = 36000

What is the minimum number of equilateral triangles, of side length 1 unit, needed to cover an equilateral triangle of side length 10 units?

The ratio of the sides of the small to big equilateral triangle (note that they are similar) is 1/10, so the ratio of their areas is (1/10)^2 = 1/100. So the big equilateral triangle has 100 times the area of a small one, so it will take {100} small triangles to cover the big one.

Centroid of a traingle

the point at which all three medians intersect. It divides each median at a ratio of 2:1

Area of a regular hexagon with side length s

√3/2 * s^2

Which of the cones below can be formed from a 252º sector of a circle of radius 10 by aligning the two straight sides? A. base radius = 6, slant =10 B. base radius = 6, height =10 C. base radius = 7, slant =10 D. base radius = 7, height =10 E. base radius = 8, slant = 10

The slant height of the cone is equal to the radius of the sector, or 10. The circumference of the base of the cone is equal to the length of the sector's arc, or 252/360*(20\π) = 14π. The radius of a circle with circumference 14π is 7. Hence the answer is {C}

A frustum of a right circular cone is formed by cutting a small cone off of the top of a larger cone. If a particular frustum has an altitude of $24$ centimeters, the area of its lower base is $225\pi$ sq cm and the area of its upper base is $25\pi$ sq cm, what is the altitude of the small cone that was cut off?

The two bases are circles, and the area of a circle is $\pi r^2$. If the area of the upper base (which is also the base of the small cone) is $25\pi$ sq cm, then its radius is $5$ cm, and the radius of the lower base is $15$ cm. The upper base, therefore, has a radius that is 1/3 the size of the radius of the smaller base. Because the slope of the sides of a cone is uniform, the frustum must have been cut off 2/3 of the way up the cone, so $x$ is 1/3 of the total height of the cone, $H$. We can now solve for $x$, because we know that the height of the frustum, $24$ cm is 2/3 of the total height, which is 36.

How many 3-letter words can we make from the letters A, B, C, and D, if we are allowed to repeat letters, and we must use the letter A at least once? (Here, a word is an arbitrary sequence of letters.)

There are $4^3$ three letter words from A, B, C, and D, and there are $3^3$ three letter words from just B, C, and D. There must, then, be $4^3 - 3^3=64-27 = {37}$ words from A, B, C, and D containing at least one A.

The inhabitants of the island of Jumble use the standard Roman alphabet (26 letters, A through Z). Each word in their language is 3 letters, and for some reason, they insist that all words contain the letter A at least once. How many 3-letter words are possible?

There are 26^3 possible words that can be formed. Of those words, 25^3 have no A. Thus, our answer is 26^3 - 25^3 = {1951}.

(Interior and Exterior Angles) A regular polygon has interior angles of 144 degrees. How many sides does the polygon have?

Use formula Interior Angle = ((Number of Sides -2) * 180)/Number of Angles Let n be the number of sides in the polygon. The sum of the interior angles in any n-sided polygon is 180(n-2) degrees. Since each angle in the given polygon measures 144º, the sum of the interior angles of this polygon is also $144n$. Therefore, we must have 180(n-2) = 144n Expanding the left side gives 180n - 360 = 144n, so 36n = 360 and n = {10}.

Two sides of a nondegenerate triangle measure 2'' and 4'' and the third side measures some whole number of inches. If a cube with faces numbered 1 through 6 is rolled, what is the probability, expressed as a common fraction, that the number showing on top could be the number of inches in the length of the third side of the triangle?

Using the Triangle Inequality, if two of the sides are 2" and 4," that means the third side must be larger than 2" but smaller than 6". That means of the six possible rolls, only three (3, 4, 5) work. Therefore, our answer is 1/2

How many integers from 1 through 9999, inclusive, do not contain any of the digits 2, 3, 4 or 5?

We have 6 digits to choose from: 0, 1, 6, 7, 8, and 9. We therefore have 6 choices for each of the digits in a 4-digit number, where we think of numbers with fewer than four digits as having leading 0s. (For example, 0097 is 97.) Since we have 6 choices for each of the four digits in the number, there are $6^4 = 1296$ ways to form the number. However, we must exclude 0000 since this is not between 1 and 9999, inclusive, so there are $1296-1 ={1295} numbers.

Altitudes AD and BE of ABC intersect at H. If angle BAC = 54º and angle ABC = 52º then what is angle AHB?

We have angle AHB = angle DHE, and from quadrilateral CDHE, we have angle DHE = 360º - angle HEC - angle ECD - angle CDH = 360º - 90º - angle ACB - 90º = 180º - angle ACB. From triangle ABC, we have 180º - angle ACB = angle BAC + angle ABC = 54º + 52º = 106º

The Gnollish language consists of 3 words, ``splargh,'' ``glumph,'' and ``amr.'' In a sentence, ``splargh'' cannot come directly before ``glumph''; all other sentences are grammatically correct (including sentences with repeated words). How many valid 3-word sentences are there in Gnollish?

We proceed by counting the complement, or the number of invalid 3-word sentences. A sentence is invalid precisely when it is of the form ``(word) splargh glumph'' or ``splargh glumph (word).'' There are 3 choices for the missing word in each sentence, and since each case is exclusive, we have a total of 6 invalid sentences. Since there are 3 * 3 * 3 = 27 possible 3-word sentences with no restrictions, there are 27-6 = {21} that satisfy the restrictions of the problem.

When finding area of part of a circle, use the sector area formula,

(θ)/2 * r^2

Area of a parallelogram

Base * height

Two chords, AB and CD, meet inside a circle at P If AP = 3 and CP = 8, then what is BP/DP?

By the POWER OF A POINT FORMULA- AP * BP = CP * DP So 3 * BP = 8 * DP and BP/DP = {8/3}

In parallelogram PQRS, the measure of angle P is five times the measure of angle Q. What is the measure of angle R, in degrees?

Consecutive angles in a parallelogram are supplementary, while opposite angles are equal. So P + Q = 180 = 5Q + Q, implying that Q = 30. Thus P = {150} = R, and we are done.

(HOW TO DEAL WITH PERCENT INCREASES) An equilateral triangle has sides of length 2 units. A second equilateral triangle is formed having sides that are $150\%$ of the length of the sides of the first triangle. A third equilateral triangle is formed having sides that are $150\%$ of the length of the sides of the second triangle. The process is continued until four equilateral triangles exist. What will be the percent increase in the perimeter from the first triangle to the fourth triangle? Express your answer to the nearest tenth.

If the side length of each successive equilateral triangle is $150\%$ of the previous triangle, then we can multiply the previous side length by 1.5. We will need to do this three times to get to the fourth triangle, so its side length will be 1.5*1.5*1.5 =3.375 times the original side length. This is the same as 337.5% of the original side length, which represents a !! 337.5 - 100 = 237.5 !! increase over the original side length. The perimeter is also a length, so it will be affected in the same way. The percent increase in the perimeter is 237.5%

A white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red stripe with a horizontal width of 3 feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet?

If the stripe were cut from the silo and spread flat, it would form a parallelogram 3 feet wide and 80 feet high. So the area of the stripe is 3(80)={240} square feet. Notice that neither the diameter of the cylinder nor the number of times the stripe wrapped around the cylinder factored into our computation for the area of the stripe. At first, this may sound counter-intuitive. An area of 240 square feet is what we would expect for a perfectly rectangular stripe that went straight up the side of the cylinder. However, note that no matter how many times the stripe is wrapped around the cylinder, its base and height (which are perpendicular) are always preserved. So, the area remains the same. Consider the following stripes which have been "unwound" from a cylinder with height 80 feet.

How many three-digit numbers are multiples of neither 5 nor 7?

It's easy to count the number of three-digit numbers which are multiples of 5 or 7: the smallest multiple of 7 which is a three-digit number is $15 \times 7 = 105$, and the largest multiple of 7 that is a three-digit number is $142 \times 7 = 994$. Therefore, there are $142-15+1 = 128$ three-digit numbers that are multiples of 7. The smallest multiple of 5 that is a three-digit number is $20\times 5 = 100$, and the largest multiple of 5 that is a three digit number is $199\times 5 =995$. So there are $199-20+1=180$ multiples of 5. Now notice that we have counted some numbers twice: those multiples of $5\times7=35$. The smallest multiple of 35 is $3\times 35 = 105$, the largest multiple of 35 is $28\times35 =980$. So there are $28-3+1=26$ multiples of 35. We have 128 multiples of 7 and 180 multiples of 5, but we count 26 multiples twice. So, there are a total of $128+180-26 = 282$ distinct three-digit numbers that are multiples of 5 or 7 (or both). There are 900 three-digit numbers in total (from 100 to 999), so there are $900-282 = {618}$ three-digit numbers that are not multiples of 7 nor 5.

Polyhedron $P$ is inscribed in a sphere of radius $36$ (meaning that all vertices of $P$ lie on the sphere surface). What is the least upper bound on the ratio (volume of P)/(surface of P)

Let $O$ be the center of the sphere, and assume for now that $O$ is inside polyhedron $P$. We can carve polyhedron $P$ into pyramids, each of which has a face of $P$ as its base and $O$ as its apex. For example, a cube would be carved into six pyramids. Then if we add up the areas of all the pyramids' bases, we get the surface area of $P$. If we add up the volumes of the pyramids, we get the volume of $P$. The volume of each pyramid is equal to $\frac 13\cdot\text{(area of base)}\cdot\text{(height)}$. The height of each pyramid must be less than $36$, since the height of each pyramid extends from $O$ to a point inside the sphere. Therefore, the volume of each pyramid is less than $12$ times the area of the base. It follows that the volume of $P$ is less than $12$ times the surface area of $P$. We can, however, make this ratio arbitrarily close to $12$ by selecting polyhedra $P$ with many small faces, so that the height of each pyramid is as close as we wish to $36$. Therefore, for polyhedra inscribed in a sphere of radius $36$ such that the center of the sphere lies inside the polyhedron, the least upper bound on the ratio of volume to surface area is 12

Side $AB$ of regular hexagon $ABCDEF$ is extended past $B$ to point $X$ such that $AX = 3AB$. Given that each side of the hexagon is $2$ units long, what is the length of segment $FX$? Express your answer in simplest radical form.

Let $P$ be the foot of the perpendicular from $F$ to the line containing $AB$. Since $\angle FAB = 120^{\circ},$ then $\angle PAF = 180^\circ - 120^\circ = 60^{\circ}$, and it follows that $\triangle PAF$ is a $30-60-90$ triangle. As $AF = 2$, it follows that $AP = 1$ and $PF = \sqrt{3}$. Also, $AB = 2$ and so $AX = 3AB = 6$. Thus, $PX = AP + AX = 7$. In right triangle $FPX$, by the Pythagorean Theorem, it follows that FX^2 = PF^2 = 7^2 +sqrt(3)^2 = 52 FX = sqrt(52) = 2*sqrt(13)

The ratio of two complementary angles is 3:2. What is the measure, in degrees, of the smaller angle?

Let the measure of the larger angle be $3x$. Then the measure of the smaller angle is $2x$, and because the angles are complementary we have 3x+2x=90. It follows that x = 90/5 =18 , so the measure of the smaller angle is 2x={36} degrees.

Let ß(x) denote the sum of the digits of the positive integer x. For example, ß(8) = 8 and ß(123) = 1+2+3=6. For how many two-digit values of x is ß(ß(x)) = 3?

Let y=ß (x). Since x ≤ 99, we have y ≤ 18. Thus if ß(y)=3, then y=3 or y=12. The 3 values of x for which ß (x)=3 are 12, 21, and 30, and the 7 values of x for which (x)=12 are 39, 48, 57, 66, 75, 84, and 93. There are {10} values in all.

The sequence 2, 3, 5, 6, 7, 10, 11, ... contains all the positive integers from least to greatest that are neither squares nor cubes. What is the 400th term of the sequence?

Let's try counting the number of perfect squares and cubes less than $441=21^2$. There are twenty perfect squares less than 441: $1^2, 2^2, \ldots, 20^2$. There are also seven perfect cubes less than 441: $1^3, 2^3, \ldots, 7^3$. So there would seem to be 20+7=27 numbers less than 441 which are either perfect squares and perfect cubes. But wait! $1=1^2=1^3$ is both a perfect square and a perfect cube, so we've accidentally counted it twice. Similarly, we've counted any sixth power less than 441 twice because any sixth power is both a square and a cube at the same time. Fortunately, the only other such one is $2^6=64$. Thus, there are 27-2=25 numbers less than 441 that are perfect squares or perfect cubes. Also, since $20^2=400$ and $7^3=343$, then all 25 of these numbers are no more than 400. To compensate for these twenty-five numbers missing from the list, we need to add the next twenty-five numbers: 401, 402, .., 424, 425, none of which are perfect square or perfect cubes. Thus, the 400th term is ${425}$.

(3D Right Triangle Problem) Triangle PAB and square ABCD are in perpendicular planes. Given that PA=3, PB=4, and AB=5, what is PD?

Since line segment $AD$ is perpendicular to the plane of $PAB$, angle $PAD$ is a right angle. In right triangle $PAD, PA=3, AD=AB=5$. By the Pythagorean Theorem PD = sqrt{3^2+5^2}={sqrt{34}}. The fact that PB=4 was not needed.